This puzzle- subset sum?

Question

In today's edition of the Guardian (a newspaper in the UK), in the "Pyrgic puzzles" section on page 43 by Chris Maslanka, the following puzzle was given:

The 3 wise men ... went to Herrods to do their Christmas shopping. Caspar bought Gold, Melchior bought Frankincense, and Balthazar bought a copy of the Daily Myrrh. The cashier tapped in the number of euros of each of these things cost, and meant to add the three numbers, but multiplied them instead. ... the marvel of the thing was that the result was exactly the same: €65.52. What were the three sums [I assume he meant the three numbers]?

My interpretation is: Find an a, b and c such that a + b + c = abc = 65.52 (exactly) where a, b and c are positive decimal numbers with no more than two decimal places. It follows that a, b and c must also be less than 65.52 (approximately).

My approach is thus: I shall find all the candidate sets of a, b and c where a + b + c = 6552 and a, b and c are integers from {1 ... 6550} (Notionally I have multiplied all the operands by 100 for convenience). Then, for all the candidate sets, it is trivial to satisfy the other condition by dividing all the operands by 100 then multiplying them (with arbitrary-precision arithmetic).

This, as I see it, is an instance of the subset sum problem. So I implemented a dirty (exponential time) algorithm which found one distinct solution: a=0.52, b=2, c=63.

Ok, there are better algorithms for the subset sum problem, but don't you think this is getting a little out-of-reach for an average Guardian reader?

On page 40 the answer is listed:

This is easy, by trial and error. Guess 52p for the Daily Myrrh. But by multiplying by 0.52 is roughly halving, so we need one sum to be about 2; so try 2 X 63 X 0.52. Et voilà. Is this answer unique?

Well, we know that the answer is unique (disregarding the other permutations of 2, 63 and 0.52).

What I want to know is: How can this be "easy"? Am I right in characterising the puzzle as an instance of the subset sum problem? Have I overlooked some characteristic of the puzzle which can be utilized to simplify the solution? Was anyone able to adopt a similar "trial and error" approach and if so can they take me through it? Is Chris Maslanka simply undaunted by NP-complete problems?

Answer 1

No, it is not an instance of the subset sum problem, because:

1. The subset size is limited to 3, making it O(n^3) solution worst case with naive exhaustive search (and not exponential)
2. There is additional data in here, the product of the numbers.
3. You are not actually given a set, a set of all integers is just a subproblem of subset-sum, a much easier one.

The important thing to understand here is: if a problem can be solved by an NP-Hard problem - it doesn't mean it is NP-Hard as well, the other way around holds - if you have a problem, and you can solve some NP-Hard problem (polynomially) with it, then your problem is NP-Hard. It is called polynomial reduction¹.

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The approach is easy because all you have to do is "guess" (by iterating all candidates) a value for a, and from this you can derive what is the possible solution for b,c - (2 variables, two equations if a is known - and in each iteration - it is), thus the solution is even linear - not only sub exponential.

It might even be optimized to use a variation of binary search to get a sub-linear optimization, but I cannot think of that optimization at the moment.

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(1) Note: this is some intuitive explanation, and not a formal definition.